Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <queue>
#define make make_pair
#define N 100002
#define INF (1e9)
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
void bfs(ll,ll);
ll n,m,s,t,a,b,cost[2][N],mincost,ans;
vector<ll> c[N];
ll C[N];

int main(){
  cin>>n>>m>>s>>t;
  for(int i=0;i<m;i++){
    cin>>a>>b;
    c[a].push_back(b);
    c[b].push_back(a);
  }
  bfs(0,s);
  bfs(1,t);
  for(int i=0;i<=mincost-2;i++) ans+=cost[0][i]*cost[1][mincost-2-i];
  if(mincost-2<0) cout<<0<<endl;
  else cout<<ans<<endl;
  return 0;
}

void bfs(ll index,ll start){
  queue<P> q;
  for(int i=1;i<=n;i++) C[i]=INF;
  q.push(make(0,start));
  C[start]=0;
  while(!q.empty()){
    P k=q.front(); q.pop();
    int cos=k.first,u=k.second;
    for(int i=0;i<(int)c[u].size();i++){
      int nu=c[u][i];
      if(C[nu]>cos+1){
	C[nu]=cos+1;
	q.push(make(C[nu],nu));
      }
    }
  }
  if(!index) mincost=C[t];
  for(int i=1;i<=n;i++){
    if(C[i]==INF) continue;
    cost[index][C[i]]++;
  }
}